I'm having trouble understanding how to find a limiting distribution. If I have a Markov Chain whose transition probability matrix is: $$ \mathbf{P} = \matrix{~ & 0 & 1 & 2 & 3 & 4 \\ 0 & q & p & 0 & 0 & 0 \\ 1 & q & 0 & p & 0 & 0 \\ 2 & q & 0 & 0 & p & 0\\ 3 & q & 0 & 0 & 0 & p \\ 4 & 1 & 0 & 0 & 0 & 0 } $$
where p>0, q>0 and p+q=1
How would I go about finding the limiting distribution? Thanks for any and all help!
On
There is no stable limiting distribution. The transition matrix is not diagonalizable.
The Markov chain has this state diagram:
Now suppose that at an arbitrarily large number of steps later, the chain is in state $0$ with probability $a$, state $1$ with probability $b$, state $2$ with probability $c$, state $3$ with probability $d$, and state $4$ with probability $e$, and suppose, for contradiction, that this is a constant limiting distribution.
Then
$$a = bp + e$$
$$b = aq$$
$$c = bq$$
$$d = cq$$
$$e = dq$$
Substitution gives $e = aq^4$, so
$$a = aqp + aq^4 \Rightarrow 1 = qp + q^4$$
We also know $p+q=1$, so
$$1 = q(1-q)+q^4$$
$$q^4 -q^2 + q - 1$$
$$q^2(q+1)(q-1)+1(q-1)=0$$
Since $q \neq 1$,
$$q^3 +q^2 +1 = 0$$
This has no positive solutions for $q$ (by inspection). Therefore, there is no stable limiting distribution.
HINT: Diagonalize the matrix $\mathbf{P}$.